Question

prove that 7-3√2 is irrational number​

Answer 1

Answer:

Step-by-step explanation:

Let us assume that  7-3√2  is a rational.

                     Then,       7-3√2 =  $\frac{a}{b}$

                             ⇒    $-3\sqrt{2} =\frac{a-7b}{b}$

                            ⇒         $\sqrt{2} =\frac{a-7b}{-3b}$

$\frac{a-7b}{-3b}$  is a rational.

√2  is also rational.

But this contradicts the fact that √2 is irrational. so our assumption was wrong . Hence  7-3√2  is irrational​.

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Answer 2

Given

A real number: $7-3\sqrt{2}$.

To Prove,

The number $7-3\sqrt{2}$ is an irrational number.

Solution,

The method of proving the number $7-3\sqrt{2}$ irrational is as follows -

We will prove it by contradiction.

If possible let us suppose that $7-3\sqrt{2}$ is a rational number.

We know that a rational number can always be represented as $\frac{x}{y}$ where x and y are two co-prime integers and $y\neq0$.

Let $7-3\sqrt{2}=\frac{p}{q}$, where p and q are two non-zero, co-prime integers.

So, $7-3\sqrt{2}=\frac{p}{q}$ ⇒  $3\sqrt{2} =7-\frac{p}{q}=\frac{7q-p}{q}$

⇒  $\sqrt{2}=\frac{7q-p}{3q}$ ...... (I)

Now we can observe that the R.H.S. of equation I is a rational number because (7q - p) and 3q are two non-zero integers. But this rational number is equal to $\sqrt{2}$ which is an irrational number.

So there is a contradiction. So our hypothesis was wrong.

Hence, it is proved that $7-3\sqrt{2}$ is an irrational number.

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